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Question

Find the sum of the following series to n terms : (1-7) 13+33+53+73+....


Solution

13+33+53+73+....Let Tn be the nth term of this series then,Tn=[1+(n1)2]3=(2n1)3=(2n)33(2n)2.1+3.12.2n13=8n312n2+6n1[(ab)3=a33a2b+3ab2b3]=nk1Tk=nk1(8k312k2+6k1)=8nk1k312nk1k2+6nk1knk11=8[n(n+1)22]12[n(n+1)(2n+1)6]+6[n(n+1)2]n=8n2(n+1)2412[n(n+1)(2n+1)]6+3[n(n+1)]n=2n2(n+1)22n(n+1)(2n+1)+3(n+1)nn=n(n+1)[2n2(n+1)2n(2n+1)+3n]n=(n+1)[2n3+2n24n22n+3n]n=(n+1)[2n32n2+n]n=2n42n3+n2+2n32n2+nn=2n4n2=n2(2n21)13+33+53+73+.... to n terms =n2(2n21)

Mathematics
RD Sharma
Standard XI

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